Question

A bullet of mass 0.01 kg fired from a gun hits a wooden plank with a velocity of 300 m/s,emerges out with a velocity of 100 m/s .calculate the average resistance offered by the plank to the motion of the bullet if the thickness of the plank is 0.05m

Answer 1

Answer:

Average resistive force is= mass (m) × a

                                           =8000N

Step-by-step explanation:

Initial velocity u =300m/s

Final velocity v= 100m/s

Distance s=0.05m

Retartdation be a

2×a×s= v²- u²

2×a×0.05= 100×100  - 300×300

a= 800000

Average resistive force is= mass (m) × a

                                           =8000N

Answer 2

The average resistance offered by the plank to the motion of bullet if the thickness of the plank is 0.05 m=-40000 N

Step-by-step explanation:

Mass of bullet=m=0.01 Kg

Initial velocity=u=300m/s

Final velocity=v=100m/s

Displacement=S=0.05 m

We know that

$v^2-u^2=2as$

$(100)^2-(300)^2=2a(0.05)$

$10000-90000=0.1a$

$-80000=0.1a$

$a=-\frac{80000}{0.1}=-800000m/s^2$

We know that

Force=$F=mass(m)\times acceleration(a)$

Using the formula

Force=$-800000\times 0.05=-40000 N$

Where negative sign indicates that the force applied in opposite direction of motion.

Therefore, the average resistance offered by the plank to the motion of bullet =-40000 N

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